path: root/frsb_kac-rice.tex

\documentclass[fleqn]{article}

\usepackage{fullpage,amsmath,amssymb,latexsym,graphicx}

\begin{document}
\title{Full solution of the Kac--Rice problem for mean-field models.\\
or Full solution for the counting of saddles of mean-field glass models}
\author{Jaron Kent-Dobias \& Jorge Kurchan}
\maketitle
\begin{abstract}
    We derive the general solution for the computation of saddle points
    of  mean-field complex landscapes. The solution incorporates Parisi's solution
    for equilibrium, as it should.
\end{abstract}
\section{Introduction}

The computation of the number of metastable states of mean field spin glasses
goes back to the beginning of the field. Over forty years ago,
Bray and Moore \cite{Bray_1980_Metastable} attempted the first calculation for 
 the Sherrington-Kirkpatrick model, a paper remarkable for being the first practical application of a replica symmetry breaking scheme. As became clear when the actual
 ground-state of the model was computed by Parisi \cite{Parisi_1979_Infinite}, the Bray-Moore result
 was not exact, and
in fact  the problem has been open 
ever since.
Indeed, to this date the program of computing the number of saddles of a mean-field 
glass has been only  carried out for a small subset of models.
These include most notably the $p$-spin model ($p>2$) \cite{Rieger_1992_The, Crisanti_1995_Thouless-Anderson-Palmer}. 
The problem of studying the critical points of these landscapes
has evolved into an active field in probability theory \cite{Auffinger_2012_Random, Auffinger_2013_Complexity, BenArous_2019_Geometry}

In this paper we present what we argue is the general replica ansatz for the 
computation of the number of saddles of generic mean-field models,  including the Sherrington-Kirkpatrick model. It incorporates  the Parisi solution as the limit of lowest states, as it should.


\section{The model}

Here we consider, for definiteness, the  mixed $p$-spin model, itself a particular case 
of the `Toy Model' of M\'ezard and Parisi \cite{Mezard_1992_Manifolds}
\begin{equation}
  H(s)=\sum_p\frac{a_p^{1/2}}{p!}\sum_{i_1\cdots i_p}J_{i_1\cdots i_p}s_{i_1}\cdots s_{i_p}
\end{equation}
for $\overline{J^2}=p!/2N^{p-1}$. Then
\begin{equation}
  \overline{H(s_1)H(s_2)}=Nf\left(\frac{s_1\cdot s_2}N\right)
\end{equation}
for
\begin{equation}
  f(q)=\frac12\sum_pa_pq^p
\end{equation}
Can be thought of as a model of generic gaussian functions on the sphere.
To constrain the model to the sphere, we use a Lagrange multiplier $\mu$, with the total energy being
\begin{equation}
  H(s)+\frac\mu2(N-s\cdot s)
\end{equation}

At any critical point, the hessian is
\begin{equation}
  \operatorname{Hess}H=\partial\partial H-\mu I
\end{equation}
$\partial\partial H$ is a GOE matrix with variance
\begin{equation}
  \overline{(\partial_i\partial_jH)^2}=\frac1Nf''(1)\delta_{ij}
\end{equation}
and therefore its spectrum is given by the Wigner semicircle with radius $\sqrt{4f''(1)}$, or
\begin{equation}
  \rho(\lambda)=\frac1{\pi\sqrt{f''(1)}}\sqrt{\lambda^2-4f''(1)}
\end{equation}
and the spectrum of $\operatorname{Hess}H$ is this shifted by $\mu$, or $\rho(\lambda-\mu)$.

The parameter $\mu$ fixes the spectrum of the hessian. By manipulating it, one
can decide to find the complexity of saddles of a certain macroscopic index, or
of minima with a certain harmonic stiffness. When $\mu$ is taken to be within
the range $\pm\sqrt{4f''(1)}=\pm\mu_m$, the critical points are constrained to have
index $\frac12N(1-\mu/\mu_m)$. When $\mu>\mu_m$, the critical
points are minima whose sloppiest eigenvalue is $\mu-\mu_m$. Finally,
when $\mu=\mu_m$, the critical points are marginal minima.


\subsection{What to expect?}

In order to try to visualize what one should expect, consider two pure p-spin models, with 
\begin{equation}
    H = H_1 + H_2=\alpha_1 \sum_{ijk} J^1_{ijk} s_i s_j s_k +
    \alpha_2 \sum_{ijk} J^2_{ijk} \bar s_i \bar s_j \bar s_k +\epsilon \sum_i s_i \bar s_i
\end{equation}
The complexity of the first and second systems in terms of $H_1$ and of $H_2$ 
have, in the absence of coupling,  the same dependence, but are stretched to one another:
\begin{equation}
    \Sigma_1(H_1)= \Sigma_o(H_1/\alpha_1) \qquad ; \qquad \Sigma_2(H_2)= \Sigma_o(H_2/\alpha_2) 
\end{equation}
Each system has a ground state energy $E_{gs}^{1,2}$, a threshold energy $E_{thres}^{1,2}$ (a well-defined notion, since we are considering pure p-spins), abd the corresponding limit values $X^{1,2}_{gs}=\left. \frac{d \Sigma_1}{dE_{1,2}}\right|_{E^{gs}_{12}}$
and $X^{1,2}_{thres}=\left. \frac{d \Sigma_1}{dE_{1,2}}\right|_{E^{thres}_{12}}$
Considering the cartesian product of both systems, we have, in terms of the total energy
$H=H_1+H_2$ three regimes:
\begin{itemize}
\item  {\bf Unfrozen}: 
\begin{eqnarray}
& & X_1 \equiv \frac{d \Sigma_1}{dE_1}=  X_2 \equiv \frac{d \Sigma_2}{dE_2}
 \end{eqnarray}
\item  {\bf Semi-frozen}
As we go down in energy, one of the systems (say, the first) reaches its ground state,
At lower temperatures, the first system is thus frozen, while the second is not,
so that $X_1=X_1^{gs}> X_2$. The lowest energy is such that both systems are frozen.
\item {\bf Semi-threshold }  As we go up from the unfrozen upwards in energy,
the second system reaches its threshold $X_2^{thres}$. At higher energies minima are extremely rare,
so the second system remains stuck at its threshold for higher energies.

\item{\bf Both systems reach their thresholds} There essentially no more minima above that.
\end{itemize}
Consider now two combined vectors $({\bf s},{\bf \hat s})$ and $({\bf s}',{\bf \hat s}')$
chosen at the same energies.\\

$\bullet$ Their normalized overlap is close to one when both subsystems are frozen,
close to a half in the semifrozen phase, and zero at all higher energies.\\

$\bullet$ In phases where one or both systems are stuck in their thresholds (and only in those), the
minima are exponentially subdominant with respect to saddles.

$\bullet$ {\bf note that the same reasoning leads us to the conclusion that
minima of two total energies such that one of the systems is frozen have nonzero overlaps}

\section{Main result}

\begin{equation}
  \begin{aligned}
    \overline{\Sigma(\epsilon,\mu)}
    =\mathcal D(\mu)
    +\operatorname*{extremum}_{\substack{R_d,D_d,\hat\epsilon\in\mathbb R\\\chi\in\Lambda}}
    \left\{
      \hat\epsilon\epsilon+\mu R_d
      +\frac12(2\hat\epsilon R_d-D_d)f'(1)+\frac12R_d^2f''(1)
      +\frac12\log R_d^2 \right.\\\left.
      +\frac12\int_0^1dq\,\left(
        \hat\epsilon^2f''(q)\chi(q)+\frac1{\chi(q)+R_d^2/D_d}
      \right)
    \right\}
  \end{aligned}
\end{equation}
where
\begin{equation}
  \mathcal D(\mu)
  =\operatorname{Re}\left\{
    \frac12\left(1+\frac\mu{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)
    -\log\left(\frac1{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)
  \right\}
\end{equation}
and $\Lambda$ is the space of functions $\chi:[0,1]\to[0,1]$ which are
monotonically decreasing, convex, and have $\chi(1)=0$ and $\chi'(1)=-1$.
If there is more than one extremum of this function, choose the one with the
smallest value of $\Sigma$. The sign of the root inside $\mathcal D(\mu)$ is
negative for $\mu>0$ and positive for $\mu<0$.

The $k$-RSB ansatz is equivalent to piecewise linear $\chi$ with $k+1$
pieces, with replica symmetric or 0-RSB giving $\chi(q)=1-q$. Our other major
result is that, if the equilibrium state in the vicinity of zero temperature is
given by a $k$-RSB ansatz, then the complexity is given by a $(k-1)$-RSB
ansatz. Moreover, there is an exact correspondence between the parameters of
the equilibrium saddle point in the limit of zero temperature and those of the
complexity saddle at the ground state. If the equilibrium is given by
$x_1,\ldots,x_k$ and $q_1,\ldots,q_k$, then the parameters $\tilde
x_1,\ldots,\tilde x_{k-1}$ and $\tilde q_1,\ldots,\tilde q_{k-1}$ for the
complexity in the ground state are
\begin{align}
  \hat\epsilon=\lim_{\beta\to\infty}\beta x_k
  &&
  \tilde x_i=\lim_{\beta\to\infty}\frac{x_i}{x_k}
  &&
  \tilde q_i=\lim_{\beta\to\infty}q_i
  &&
  R_d=\lim_{\beta\to\infty}\beta(1-q_k)
  &&
  D_d=R_d\hat\epsilon
\end{align}

\section{Equilibrium}

Here we review the equilibrium solution. \cite{Crisanti_1992_The, Crisanti_1993_The, Crisanti_2004_Spherical, Crisanti_2006_Spherical}

\begin{equation}
  \beta F=-\frac12\lim_{n\to0}\frac1n\left(\beta^2\sum_{ab}f(Q_{ab})+\log\det Q\right)-\frac12\log S_\infty
\end{equation}
$\log S_\infty=1+\log2\pi$.
\begin{align*}
  \beta F=
  -\frac12\log S_\infty
  -\frac12\lim_{n\to0}\frac1n\left(\beta^2nf(1)+\beta^2\sum_{i=0}^kn(x_i-x_{i+1})f(q_i)
  +\log\left[
    \frac{
      1+\sum_{i=0}^k(x_i-x_{i+1})q_i
    }{
      1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0
    }
  \right]\right.\\
  +\frac n{x_1}\log\left[
      1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0
  \right]\\
  \left.+\sum_{j=1}^kn(x_{j+1}^{-1}-x_j^{-1})\log\left[
      1+\sum_{i=j+1}^k(x_i-x_{i+1})q_i-x_{j+1}q_j
  \right]
\right)
\end{align*}

\begin{align*}
  \lim_{n\to0}\frac1n
  \log\left[
    \frac{
      1+\sum_{i=0}^k(x_i-x_{i+1})q_i
    }{
      1+\sum_{i=1}^k(x_i-x_{i+1})q_i-x_1q_0
    }
  \right]
  &=
  \lim_{n\to0}\frac1n
  \log\left[
    \frac{
      1+\sum_{i=0}^k(x_i-x_{i+1})q_i
    }{
      1+\sum_{i=0}^k(x_i-x_{i+1})q_i-nq_0
    }
  \right] \\
  &=q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1}
\end{align*}


\begin{align*}
  \beta F=
  -\frac12\log S_\infty
  -\frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i)
  +q_0\left(1+\sum_{i=0}^k(x_i-x_{i+1})q_i\right)^{-1}\right. \\
  +\frac1{x_1}\log\left[
      1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i-x_1q_0
  \right]\\
  \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[
      1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j
  \right]
\right)
\end{align*}
$q_0=0$
\begin{align*}
  \beta F=
  -\frac12\log S_\infty
  -\frac12\left(\beta^2f(1)+\beta^2\sum_{i=0}^k(x_i-x_{i+1})f(q_i)
  +\frac1{x_1}\log\left[
      1+\sum_{i=1}^{k}(x_i-x_{i+1})q_i
  \right]\right.\\
  \left.+\sum_{j=1}^k(x_{j+1}^{-1}-x_j^{-1})\log\left[
      1+\sum_{i=j+1}^{k}(x_i-x_{i+1})q_i-x_{j+1}q_j
  \right]
\right)
\end{align*}
$x_i=\tilde x_ix_k$, $x_k=y/\beta$, $q_k=1-z/\beta$
\begin{align*}
  \beta F=
  -\frac12\log S_\infty-
  \frac12\left(\beta^2f(1)+\beta^2(y\beta^{-1}-1)f(1-z\beta^{-1})+y\beta\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i)\right. \\
  +\frac\beta{\tilde x_1 y}\log\left[
    y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta
  \right]\\
  +\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[
    y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-yz/\beta-y\tilde x_{j+1}q_j
\right]\\
    \left.-\frac\beta{\tilde x_1 y}\log\beta-\sum_{j=1}^{k-1}\frac\beta y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\beta+(1-\beta y^{-1})\log\left[
      z/\beta
  \right]
\right)
\end{align*}
\begin{align*}
  \lim_{\beta\to\infty}F=
  -\frac12\left(yf(1)+zf'(1)+y\sum_{i=0}^{k-1}(\tilde x_i-\tilde x_{i+1})f(q_i)
  +\frac1{\tilde x_1 y}\log\left[
    y\sum_{i=1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z
  \right]\right.\\
  \left.+\sum_{j=1}^{k-1}\frac1 y(\tilde x_{j+1}^{-1}-\tilde x_j^{-1})\log\left[
    y\sum_{i=j+1}^{k-1}(\tilde x_i-\tilde x_{i+1})q_i+y+z-y\tilde x_{j+1}q_j
    \right]
    -\frac1y\log z
\right)
\end{align*}
$F$ is a $k-1$ RSB ansatz with all eigenvalues scaled by $y$ and shifted by $z$. $\tilde x_0=0$ and $\tilde x_k=1$.
\begin{equation} \label{eq:ground.state.free.energy}
  \lim_{\beta\to\infty}F=-\lim_{n\to0}\frac1n\frac12\left(nzf'(1)+y\sum_{ab}f(\tilde Q_{ab})+\frac1y\log\det(yz^{-1}\tilde Q+I)
  \right)
\end{equation}




\section{Kac--Rice}

\cite{Auffinger_2012_Random, BenArous_2019_Geometry}

\begin{equation}
  \mathcal N(\epsilon, \mu)
  =\int ds\,\delta(N\epsilon-H(s))\delta(\partial H(s)-\mu s)|\det(\partial\partial H(s)-\mu I)|
\end{equation}
\begin{equation}
  \Sigma(\epsilon,\mu)=\frac1N\log\mathcal N(\epsilon, \mu)
\end{equation}

{\em The `mass' term $\mu$ may take a fixed value, or it may be an integration constant,
for example fixing the spherical constraint.
This will turn out to be important when we discriminate between counting all solutions, or selecting those of a given index, for example minima}


\subsection{The replicated  problem}

\cite{Ros_2019_Complex}
\cite{Folena_2020_Rethinking}

\begin{equation}
  \begin{aligned}
    \Sigma(\epsilon, \mu)
    &=\frac1N\lim_{n\to0}\frac\partial{\partial n}\mathcal N^n(\epsilon) \\
    &=\frac1N\lim_{n\to0}\frac\partial{\partial n}\int\prod_a^n ds_a\,\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)-\mu s_a)|\det(\partial\partial H(s_a)-\mu I)|
  \end{aligned}
\end{equation}



\begin{equation}
  \begin{aligned}
    \overline{\Sigma(\epsilon, \mu)}
    &=\frac1N\lim_{n\to0}\frac\partial{\partial n}\int\left(\prod_a^nds_a\right)\,\overline{\prod_a^n \delta(N\epsilon-H(s_a))\delta(\partial H(s_a)-\mu s_a)}
    \times
    \overline{\prod_a^n |\det(\partial\partial H(s_a)-\mu I)|}
  \end{aligned}
\end{equation}{\bf
As noted by Bray and Dean  \cite{Bray_2007_Statistics}, gradient and Hessian are independent
for a Gaussian distribution, and
the average over disorder breaks into a product of two independent averages, one for the gradient factor and one for the determinant. The integration of all variables, including the disorder in the last factor, may be restricted to the domain such that the matrix $\partial\partial H(s_a)-\mu I$ has  a specified number of negative eigenvalues (the index {\cal{I}} of the saddle), 
(see Fyodorov \cite{Fyodorov_2007_Replica} for a detailed discussion) }

\begin{equation}
  \begin{aligned}
    \mathcal D(\mu)
    &=\frac1N\overline{\log|\det(\partial\partial H(s_a)-\mu I)|}
    =\int d\lambda\,\rho(\lambda-\mu)\log|\lambda| \\
    &=\operatorname{Re}\left\{\frac12\left(1+\frac\mu{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)-\log\left(\frac1{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)\right\}
  \end{aligned}
\end{equation}
for $\rho$ a semicircle distribution with radius $\sqrt{4f''(1)}$.

all saddles versus only minima

\begin{equation}
  \prod_a^n\delta(N\epsilon-H(s_a))\delta(\partial H(s_a)-\mu s_a)
  =\int \frac{d\hat\epsilon}{2\pi}\prod_a^n\frac{d\hat s_a}{2\pi}
    e^{\hat\epsilon(N\epsilon-H(s_a))+i\hat s_a\cdot(\partial H(s_a)-\mu s_a)}
\end{equation}

\begin{equation}
  \begin{aligned}
    \overline{
      \exp\left\{
        \sum_a^n(i\hat s_a\cdot\partial_a-\hat\epsilon)H(s_a)
      \right\}
    }
    &=\exp\left\{
        \frac12\sum_{ab}^n
        (i\hat s_a\cdot\partial_a-\hat\epsilon)
        (i\hat s_b\cdot\partial_b-\hat\epsilon)
        \overline{H(s_a)H(s_b)}
      \right\} \\
    &=\exp\left\{
        \frac N2\sum_{ab}^n
        (i\hat s_a\cdot\partial_a-\hat\epsilon)
        (i\hat s_b\cdot\partial_b-\hat\epsilon)
        f\left(\frac{s_a\cdot s_b}N\right)
      \right\} \\
    &\hspace{-13em}\exp\left\{
        \frac N2\sum_{ab}^n
        \left[
          \hat\epsilon^2f\left(\frac{s_a\cdot s_b}N\right)
          -2i\hat\epsilon\frac{\hat s_a\cdot s_b}Nf'\left(\frac{s_a\cdot s_b}N\right)
          -\frac{\hat s_a\cdot \hat s_b}Nf'\left(\frac{s_a\cdot s_b}N\right)
          +\left(i\frac{\hat s_a\cdot s_b}N\right)^2f''\left(\frac{s_a\cdot s_b}N\right)
        \right]
      \right\}
  \end{aligned}
\end{equation}

The parameters:
\begin{align}
  Q_{ab}=\frac1Ns_a\cdot s_b &&
  R_{ab}=-i\frac1N\hat s_a\cdot s_b &&
  D_{ab}=\frac1N\hat s_a\cdot\hat s_b
\end{align}

\begin{equation}
  \begin{aligned}
    S
    =\mathcal D(\mu)+\hat\epsilon\epsilon+\lim_{n\to0}\frac1n\left(
      \mu\sum_a^nR_{aa}
      +\frac12\sum_{ab}\left[
        \hat\epsilon^2f(Q_{ab})+2\hat\epsilon R_{ab}f'(Q_{ab})
        -D_{ab}f'(Q_{ab})+R_{ab}^2f''(Q_{ab})
      \right] \right. \\ \left.
    +\frac12\log\det\begin{bmatrix}Q&iR\\iR&D\end{bmatrix}
    \right)
  \end{aligned}
\end{equation}


\section{Replicated action}


\begin{equation}
  \begin{aligned}
    S
    =\mathcal D(\mu)+\hat\epsilon\epsilon+\lim_{n\to0}\frac1n\left(
      \mu\sum_a^nR_{aa}
      +\frac12\sum_{ab}\left[
        \hat\epsilon^2f(Q_{ab})+2\hat\epsilon R_{ab}f'(Q_{ab})
        -D_{ab}f'(Q_{ab})+R_{ab}^2f''(Q_{ab})
      \right] \right. \\ \left.
    +\frac12\log\det\begin{bmatrix}Q&iR\\iR&D\end{bmatrix}
    \right)
  \end{aligned}
\end{equation}
\begin{align}
  0&=\frac{\partial S}{\partial R_{ab}}
  =\mu\delta_{ab}+\hat\epsilon f'(Q_{ab})+R_{ab}f''(Q_{ab})+\sum_c(DQ+R^2)^{-1}_{ac}R_{cb} \\
  0&=\frac{\partial S}{\partial D_{ab}}
  =-\frac12 f'(Q_{ab})+\frac12\sum_c(DQ+R^2)^{-1}_{ac}Q_{cb}
\end{align}
The second equation implies
\begin{equation}
  (DQ+R^2)^{-1}=Q^{-1}f'(Q)
\end{equation}

\section{Replica ansatz}

\subsection{Motivation}

We shall make the following ansatz
to putting:
\begin{align}\label{ansatz}
  Q_{ab}= \text{a Parisi matrix} &&
  R_{ab}=R_d \delta_{ab} &&
  D_{ab}= D_d \delta_{ab}
\end{align}
This ansatz closes under the operations that are involved in the replicated action.
The reader who is happy with the ansatz may skip the rest of this section.

 We may encode the original variables in a superspace variable:
\begin{equation}
  \phi_a(1)= s_a + \bar\eta_a\theta_1+\bar\theta_1\eta_a +  \hat s_a \bar \theta_1 \theta_1
\end{equation}
\begin{equation}
\begin{aligned}
  \mathbb Q_{a,b}(1,2)&=\frac 1 N \phi_a(1)\cdot\phi_b (2) =
Q_{ab}  -i\left[\bar\theta_1\theta_1+\bar\theta_2\theta_2\right]  R_{ab}
 +(\bar\theta_1\theta_2+\theta_1\bar\theta_2)F_{ab}
 + \bar\theta_1\theta_1 \bar \theta_2 \theta_2 D_{ab} \\ 
&+  \text{odd terms in the $\bar \theta,\theta$}~.
\end{aligned}
\label{Q12}
\end{equation}
\begin{equation}
  \overline{\Sigma(\epsilon,\mu)}
  =\hat\epsilon\epsilon\lim_{n\to0}\frac1n\left[
    \mu\int d1\sum_a^n\mathbb Q_{aa}(1,1)
    +\int d2\,d1\,\frac12\sum_{ab}^n(1+\hat\epsilon\bar\theta_1\theta_1)f(\mathbb Q_{ab}(1,2))(1+\hat\epsilon\bar\theta_2\theta_2)
    +\frac12\operatorname{sdet}\mathbb Q
  \right]
\end{equation}

Here $\theta_a$, $\bar \theta_a$ are Grassmann variables, and we denote the full set of coordinates 
in a  compact form as
$1=  \theta_1 \overline\theta_1$, $d1=   d\theta_1 d\overline\theta_1$, etc.
The odd and even fermion numbers decouple, so we can neglect all odd terms in $\theta,\bar{\theta}$.

\cite{Annibale_2004_Coexistence}


The variables $\bar \theta \theta$ and $\bar \theta ' \theta'$ play
the role of `times' in a superspace treatment. We have a long experience of
making an ansatz for replicated quantum problems, which naturally involve a (Matsubara) time. The analogy strongly
suggests that only the diagonal ${\bf Q}_{aa}$ depend on the $\theta$'s. This boils down the ansatz \ref{ansatz}

Not surprisingly, this ansatz closes, as we shall see. That it closes under Hadamard products is simple.

\begin{equation}
  \begin{aligned}
  \int d3\,\mathbb Q_1(1,3)\mathbb Q_2(3,2)
  =\int d3\,(
  Q_1  -i(\bar\theta_1\theta_1+\bar\theta_3\theta_3)  R_1
   +(\bar\theta_1\theta_3+\theta_1\bar\theta_3)F_1
 + \bar\theta_1\theta_1 \bar \theta_3 \theta_3 D_1
 ) \\ (
  Q_2  -i(\bar\theta_3\theta_3+\bar\theta_2\theta_2)  R_2
   +(\bar\theta_3\theta_2+\theta_3\bar\theta_2)F_2
 + \bar\theta_3\theta_3 \bar \theta_2 \theta_2 D_2
 ) \\
 =-i(Q_1R_2+R_1Q_2)
 +Q_1D_2\bar\theta_2\theta_2+D_1Q_2\bar\theta_1\theta_1
 -i\bar\theta_1\theta_1\bar\theta_2\theta_2R_1D_2
 -i\bar\theta_1\theta_1\bar\theta_2\theta_2D_1R_2
 \end{aligned}
\end{equation}

\subsection{Solution}


Insert the diagonal ansatz $R=R_dI$, $D=D_dI$. Then
\[
  0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_d(R_d^2I-D_dQ)^{-1}
  =(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+R_dQ^{-1}f'(Q)
\]
and
\[
  Q^{-1}f'(Q)=(I+D_df'(Q))/R_d^2
\]
Substituting the second into the first, we have
\[
  0=(R_df''(1)-\mu)I+\hat\epsilon f'(Q)+\frac1{R_d}(I+D_df'(Q))
\]
\[
  0=(R_df''(1)-\mu+R_d^{-1})I+(\hat\epsilon+D_d/R_d)f'(Q)
\]
The only way for this equation to be satisfied off the diagonal for nontrivial $Q$ is for $D_d=-R_d\hat\epsilon$. We therefore have
\begin{align*}
  \Sigma
  =-\epsilon\hat\epsilon+\lim_{n\to0}\frac1n\left(
    n\mu(F_d-R_d)+\frac12n\left[
      \hat\epsilon R_df'(1)+R_d^2f''(1)-F_d^2f''(1)
    \right]
    +\frac12\sum_{ab}
      \hat\epsilon^2f(Q_{ab})
    ]\right.\\\left.
    +\frac12\log\det(\hat\epsilon R_d^{-1} Q+I)
    +n\log R_d
  -n\log F_d
  \right)
\end{align*}
Taking the saddle with respect to $\mu$ and $F_d$ yields
\[
  F_d=R_d
\]
\[
  \mu=R_d^{-1}(1+R_d^2f''(1))
\]
and gives
\begin{align*}
  \Sigma
  =\epsilon\hat\epsilon+\hat\epsilon R_d f'(1)+\frac12D_df'(1)+\lim_{n\to0}\frac1n\frac12\left(
    \hat\epsilon^2\sum_{ab}
      f(Q_{ab})
    +\log\det(-D_dR_d^{-2} Q+I)
  \right)
\end{align*}
Finally, setting $0=\Sigma$ gives
\[
  \epsilon
  =\lim_{n\to0}\frac1n\frac12\left(nR_df'(1)+\hat\epsilon\sum_{ab}
      f(Q_{ab})
      +\frac1{\hat\epsilon}\log\det(\hat\epsilon R_d^{-1} Q+I)
  \right)
\]
which is precisely \eqref{eq:ground.state.free.energy} with $R_d=z$ and $\hat\epsilon=y$. 

{\em Therefore, a $(k-1)$-RSB ansatz in Kac--Rice will predict the correct ground state energy for a model whose equilibrium state at small temperatures is $k$-RSB.}

\subsection{Full}

\begin{align*}
  \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I)
  =x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\int_{q_0^+}^{q_{k-1}}dq\,\mu(q)\log\left[\hat\epsilon R_d^{-1}\lambda(q)+1\right]
\end{align*}
where
\[
  \mu(q)=\frac{\partial x^{-1}(q)}{\partial q}
\]
Integrating by parts,
\begin{align*}
  \lim_{n\to0}\frac1n\log\det(\hat\epsilon R_d^{-1} Q+I)
  &=x_1^{-1}\log\left(\hat\epsilon R_d^{-1}(1-\bar q_k)+1\right)+\left[x^{-1}(q)\log[\hat\epsilon R_d^{-1}\lambda(q)+1]\right]_{q=q_0^+}^{q=q_{k-1}}-\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac{\lambda'(q)}{x(q)}\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1}\\
  &=\log[\hat\epsilon R_d^{-1}\lambda(q_{k-1})+1]+\frac{\hat\epsilon}{R_d}\int_{q_0^+}^{q_{k-1}}dq\,\frac1{\hat\epsilon R_d^{-1}\lambda(q)+1}
\end{align*}

\begin{align*}
  \Sigma
  =-\epsilon\hat\epsilon+
    \frac12\hat\epsilon R_df'(1)
    +\frac12\int_0^1dq\,\left[
      \hat\epsilon^2\lambda(q)f''(q)
      +\frac1{\lambda(q)+R_d/\hat\epsilon}
    \right]
\end{align*}
for $\lambda$ concave, monotonic,  $\lambda(1)=0$, and $\lambda'(1)=-1$
\[
  0=\frac{\delta\Sigma}{\delta\lambda(q)}=\frac12\hat\epsilon^2f''(q)-\frac12\frac1{(\lambda(q)+R_d/\hat\epsilon)^2}
\]
\[
  \lambda^*(q)=\frac1{\hat\epsilon}\left[f''(q)^{-1/2}-R_d\right]
\]

We suppose that solutions are given by
\begin{equation}
  \lambda(q)=\begin{cases}
    \lambda^*(q) & q<q^* \\
    1-q          & q\geq q^*
  \end{cases}
\end{equation}
where $1-q$ guarantees the boundary conditions at $q=1$, and corresponds to the 0RSB or annealed solutions (annealed Kac--Rice is recovered by substituting in $1-q$ for $\lambda$). We will need to require that $1-q^*=\lambda^*(q^*)$, i.e., continuity.

Inserting this into the complexity, we find
\begin{align*}
  \Sigma
  &=-\epsilon\hat\epsilon+\frac12\hat\epsilon R_df'(1)
  +\frac12\int_0^{q^*}dq\left[
    \hat\epsilon(f''(q)^{-1/2}-R_d)f''(q)+\hat\epsilon f''(q)^{1/2}
  \right]
  +\frac12\int_{q^*}^1dq\left[
    \hat\epsilon^2(1-q)f''(q)+\frac1{q-1+R_d/\hat\epsilon}
  \right] \\
  &=-\epsilon\hat\epsilon+\frac12\hat\epsilon R_d\left[f'(1)-f'(q^*)\right]
  +\hat\epsilon\int_0^{q^*}dq\,f''(q)^{1/2}
  +\frac12\hat\epsilon^2\int_{q^*}^1dq\,
    (1-q)f''(q)
    -\log\left[1-(1-q^*)\hat\epsilon/R_d\right]
\end{align*}
$R_d$ can be extremized now, with
\[
  R_d=\frac12\left(
    (1-q^*)\hat\epsilon\pm\sqrt{
      (1-q^*)\left(
        (1-q^*)\hat\epsilon^2+8/[f'(1)-f'(q^*)]
      \right)
    }
  \right)
\]

This all is for $\mu=\mu^*$, which counts the dominant saddles. We can also count by fixed macroscopic index $\mu$ by leaving it unoptimized in the complexity. This gives
\[
  F_d=\frac1{2f''(1)}\left[\mu\pm\sqrt{\mu^2-4f''(1)}\right]
\]
and
\begin{align*}
  \Sigma
  =-\epsilon\hat\epsilon+
    \frac12\hat\epsilon R_df'(1)
    +\frac12\int_0^1dq\,\left[
      \hat\epsilon^2\lambda(q)f''(q)
      +\frac1{\lambda(q)+R_d/\hat\epsilon}
    \right]-\mu R_d+\frac12R_d^2f''(1)+\log R_d\\
    +\operatorname{Re}\left\{\frac12\left(1+\frac\mu{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)-\log\left(\frac1{2f''(1)}\left(\mu\pm\sqrt{\mu^2-4f''(1)}\right)\right)\right\}
\end{align*}


\section{Ultrametricity rediscovered}

Three states chosen at the same energy share some common information if there is some `frozen' element common to all. Suppose we choose randomly 
these states but restrict to those whose overlaps
take values $Q_{12}$ and $Q_{13}$. Unlike an equilibrium situation, where the Gibbs measure allows us to find such pairs (in a FRSB case) the cost in probability of this in the present case will be exponential.
Once conditioned this way, we compute $Q_{23}= \min(Q_{12},Q_{13})$



{\tiny wild guess:
Consider two states chosen at  different energies, corresponding to maximal values $q^{max}_1$, $q^{max}_2$. Their mutual overlap 
should be (I am guessing this) $Q_{12}=\min(q^{max}_1, q^{max}_2)$

}
\section{Conclusion}

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